- Joined
- Oct 24, 2009
One thing that should be mentioned is that it might take less time to reach the max height than the time it takes to come down back to the ice. So maybe it take 0.10 second to reach the max height and 0.37 seconds to come down?
The actual height is still 9.6 inches regardless of the horizonal distance.
The "v0" in your formula is the verticle component of the initial velocity. If v1 is the horizontal component, then the distance is simply d = v1t. If you want to maximize your horizontal distance, aim for a 45 degree angle of take-off.
No, it takes exactly the same.
Impressions looking at a photograph, or memories of competitions past, are not measurements. They may call into question whether the 0.45 seconds is correct, but they are not measurements.
Exact measurements are not needed.
I knew I would regret it!
Your example does not apply to what the skater is doing. The skater controls Vx and Vy seperately. The takeoff angle is given by ArcTan(Vy / Vx).
Starting with Vx and Vy at the instant of takeoff:
The height of the jump will be 0.5 * Vy ^ 2 / g, where g is the acceleration due to gravity.
The distance traveled will be 2 * Vx * Vy / g
To get maximum distance for a given Vx, the skater must generate a Vy = Vx, but Vx is an independant parameter controlled by the skater.
The Vx is determined by taking the integral of the force applied to the ice with respect to time divided by the mass of the skater, taking into account the fact the impulse to the ice is an inelastic process and some energy is lost compressing anf fraturing the ice, etc.
Flatt Fan said:Exactly. In this case, Vx = Vo * cos(45), Vy = Vo * sin(45)
So if she's jumping at 60 degree angel or 90 degree angle, her Vy would increase, thus increase the height and decrease the horizontal distance, no?
You fail the physics lab. You know she jumps higher than 10 inches because you know she jumps higher than 10 inches!
Yes...but...in this problem the independently measured variable is the total time,"T," not V0 = |V|. "T" is determined solely by the vertical component (T = 2Vy/a). For fixed "T" (say, t = 0.45), the maximum height H is the same no matter what the angle is: H = h(T/2) = h(Vy/a) = Vy^2/2a. (Somethoing like that. )
The 45 degree angle comes in when you want to maximize the distance for fixed V0 (but you knew that. )
This should be a simple question.
Do you believe Mao is only 9.6 in off the ice?
I don't.
Btw, I was responding to Mathman who asserted that the height is the same regardless of distance travel. It is not.
Can you explain the sporting term "explosive jumper" that we hear in basketball and track? Even in skating if Scott is announcing
Is it possible for one person to not only jump higher but also jump faster, to accelerate into the air at a faster speed than another?
I saw Charles Barkely do this many, many times......at least it appeared he was doing it.
If this was true, isn't it possible for a figure skater to have an explosive jump - and to jump up at a faster rate, than the speed they will descend back to the ice with?
I don't mean the DrJ move you accidentily gave Jordan credit for
Is it really impossible ?
Yes. it means that the jumper has superior take-off velocity. Like the muzzle velocity of a high powered rifle compared to that of a pop-gun.
If a jumper jumps "faster" -- that is, with a higher initial velocity -- then he will automatically jump higher, no question.
Once a jumper's feet leaves the ground, however, he cannot accelerate (increase his speed while in the air). Our friend Isaac tells us why: Force = mass times acceleration. Once the jumper leaves the floor, no other force acts on him except gravity (neglecting minor effects like wind resistane.)
I think the main reason for that effect is Charles' size and fierceness. Like a charging water buffalo, it just seems like he's coming on like a freight train.
(But I have to admit that Dorothy Hamill looks like she defies the laws of physics on her delayed Axel jump. )
.
By the way, this is why the CoP gives extra bonus points on spins when a skater can accelerate when going from one position to another. It is really hard. Skaters can accomplish it by manipulating their angular momentum.
Something is lost in translation with the "professor" whose "serious" analysis on a skater totally disregard the take off speed, the distance cover, and only gives the total time. Yeah, color me skeptical.
Abstract. In modern figure skating competitions, a successful jump has become much more important factor in evaluation of the competitor's performance. World-class top-level skaters always perform various kinds of jumps with 3 and a half or 4 times of rotation of the body and they often combine two or more jumps as one sequence. Both higher velocity of rotation of the skater's body and enough airborne time after take off are essential for the successful jump with multiple rotation of the body. A higher velocity of rotation will make possible to rotate the skater's body in a given time duration ( airborne time ) and a longer duration of time in the air will also have an advantage for the multiple rotation of the body in the air in a given velocity of the rotation of the body at take off. For the higher rotation velocity in the air it is necessary to obtain the greater angular momentum of the body during take off motion and the longer duration of the airborne time is led by the higher vertical component of the initial velocity of the body at take off. However it is difficult in general to achieve both these two factors simultaneously in the take off motion.
In the presentation of the symposium, we will discuss the results of the investigation of the kinematics aspects of the jump motion in figure skating in 1998 Nagano Winter Olympic Games.
I have no idea how high Mao is off the ice and wouldn't presume to say without making the measurement myself. Is 9.6 inches wrong? I will find out for myself in Vancouver. BoP seems to know for sure, but I don't. I don't possess his highly calibrated knees, nor his ability to exactly measure angles and distances by eyeball only.